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Part of JME-10 — Thermal Properties: Expansion, Calorimetry & Heat Transfer

Heat Conduction

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Fourier's law: dQ/dt=kAΔT/LdQ/dt = kA\Delta T/L for steady-state conduction through a uniform slab. Thermal conductivity kk (W m1^{-1} K1^{-1}) ranges from silver (429) to styrofoam (0.01). The thermal resistance analogy maps directly to electrical circuits: R=L/(kA)R = L/(kA), dQ/dt=ΔT/RdQ/dt = \Delta T/R.

For composite walls in series: Rtotal=R1+R2+...R_{\text{total}} = R_1 + R_2 + ..., same heat flow through all layers. The interface temperature: Ti=Thot(dQ/dt)×R1T_i = T_{\text{hot}} - (dQ/dt) \times R_1. For parallel paths: $1/R_{\text{total}} = 1/R_1 + 1/R_2$, same temperature difference across all paths.

Effective conductivity formulas: Series (equal thicknesses): keff=2k1k2/(k1+k2)k_{\text{eff}} = 2k_1k_2/(k_1+k_2) (harmonic mean). Parallel (equal areas): keff=(k1+k2)/2k_{\text{eff}} = (k_1+k_2)/2 (arithmetic mean). In steady state with no lateral losses, temperature varies linearly along a uniform rod.

A thin insulation layer can be remarkably effective — its low kk creates high resistance despite small thickness. This is why fiberglass insulation (a few centimeters) dramatically reduces building heat loss.$

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